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hw70.sol[1]

# hw70.sol[1] - IE 335 Operations Research Optimization...

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IE 335 Operations Research - Optimization Solutions to Homework 7 Fall 2011 Problem 30 (a) Let x 1 = number of cars produced each day x 2 = number of trucks produced each day The objective function represents the the profit each day. The main constraints can be interpreted as the paint shop and body shop capacity each day, respectively. (b) min v 1 + v 2 minimze the net cost s.t. 1 40 v 1 + 1 50 v 2 300 the combination of resources needed for one car 1 50 v 1 + 1 50 v 2 200 the combination of resources needed for one truck v 1 , v 2 0 (c) Let v 1 = the fair price for using the paint shop each day v 2 = the fair price for using the body shop each day The constraint (1) in primal indicates the paint shop capacity each day, so the corresponding dual variable v 1 means the fair price you have to pay for using the paint shop each day. The same interpretation applies for constraint (2), which indicates the body shop capacity each day, so the corresponding dual variable v 2 means the fair price you have to pay for using the body shop each day. The objective function in primal maximizes the total profit each day, so the dual objective corresponds to minimize the price you have to pay for using paint shop and body shop each day. 1

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Problem 31 (a) Dual problem: min 5 v 1 + 8 v 2 (1) s.t. v 1 + 2 v 2 3 (2) 2 v 1 + 3 v 2 4 (3) v 1 + v 2 1 (4) 2 v 1 + 3 v 2 5 (5) v 1 , v 2 0 (6) (b) Graphical Solution: v 1 v 2 Graphical Solution to Problem 31 - Dual 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 v 1 + 2v 2 = 3 2v 1 + 3v 2 = 4 v 1 + v 2 = 1 2v 1 + 3v 2 = 5 5v 1 + 8v 2 = 20 5v 1 + 8v 2 = 16 v * = (1,1) The optimal solution is ( v * 1 .v * 2 ) = (1 , 1), therefore the optimal value = 5 × 1 + 8 × 1 = 13. (c) Plug in the optimal solution ( v * 1 .v * 2 ) = (1 , 1) back to constraints (2)–(5), we can see that constraints (2) and (5) are active at (1,1) but constraints (3) and (4) are inactive, which means that the corresponding primal variables x 2 and x 3 are zero since you can think the dual of the dual goes back to primal. As we know, x 2 and x 3 are zero, the primal reduces to a two dimensional problem with only unknowns x 1 and x 4 , which can be solved graphically max 3 x 1 + 5 x 4 s.t. x 1 + 2 x 4 5 2 x 1 + 3 x 4 8 x 1 , x 4 0 (d) Graphical Solution: 2
x 1 x 4 Graphical Solution to Problem 31 - Primal 0 0.5 1 1.5 2 2.5 3 3.5 4 0 0.5 1 1.5 2 2.5 3 x 1 + 2x 4 = 5 2x 1 + 3x 4 = 8 3x 1 + 5x 4 = 10 3x 1 + 5x 4 = 5 x * = (1,2) Optimum occurs at ( x * 1 .x * 4 ) = (1 , 2). Objective function value = 3 × 1 + 5 × 2 = 13, which matches the dual optimal solution. Problem 32 7-2.(a) x i = batches for process i , for i = 1 , 2 , 3 Objective : minimize the total cost Constraint 1 : the demand for at least 50 batches new product Constraint 2 : constraint of 75 tons ingredient 1 Constraint 3 : constraint of 60 tons ingredient 2 7-20.(a) The marginal cost of production is the optimal value of the dual variable for the demand c 1 = \$7556.

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hw70.sol[1] - IE 335 Operations Research Optimization...

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